A line through A(-5, -4) meets the lines x + | vedclass

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(A) Let the line passing through $A(-5, -4)$ make an angle $	heta$ with the $x$-axis. The equation of the line is $\frac{x + 5}{\cos 	heta} = \frac{

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$2x + 3y + 22 = 0$

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(A) Let the line passing through $A(-5, -4)$ make an angle $	heta$ with the $x$-axis. The equation of the line is $\frac{x + 5}{\cos 	heta} = \frac{y + 4}{\sin 	heta} = r$.For point $B$ on $x + 3y + 2 = 0$,we have $(r_1 \cos 	heta - 5) + 3(r_1 \sin 	heta - 4) + 2 = 0$,which gives $r_1(\cos 	heta + 3 \sin 	heta) = 15$,so $\frac{15}{AB} = \cos 	heta + 3 \sin 	heta$.For point $C$ on $2x + y + 4 = 0$,we have $2(r_2 \cos 	heta - 5) + (r_2 \sin 	heta - 4) + 4 = 0$,which gives $r_2(2 \cos 	heta + \sin 	heta) = 10$,so $\frac{10}{AC} = 2 \cos 	heta + \sin 	heta$.For point $D$ on $x - y - 5 = 0$,we have $(r_3 \cos 	heta - 5) - (r_3 \sin 	heta - 4) - 5 = 0$,which gives $r_3(\cos 	heta - \sin 	heta) = 6$,so $\frac{6}{AD} = \cos 	heta - \sin 	heta$.Substituting these into the given relation $(\frac{15}{AB})^2 + (\frac{10}{AC})^2 = (\frac{6}{AD})^2$,we get $(\cos 	heta + 3 \sin 	heta)^2 + (2 \cos 	heta + \sin 	heta)^2 = (\cos 	heta - \sin 	heta)^2$.Expanding this,$\cos^2 	heta + 9 \sin^2 	heta + 6 \sin 	heta \cos 	heta + 4 \cos^2 	heta + \sin^2 	heta + 4 \sin 	heta \cos 	heta = \cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta$.Simplifying,$4 \cos^2 	heta + 9 \sin^2 	heta + 12 \sin 	heta \cos 	heta = 0$,which is $(2 \cos 	heta + 3 \sin 	heta)^2 = 0$.Thus,$2 \cos 	heta + 3 \sin 	heta = 0$,so $	an 	heta = -\frac{2}{3}$.The equation of the line is $y + 4 = -\frac{2}{3}(x + 5)$,which simplifies to $2x + 3y + 22 = 0$.

$A$ line through $A(-5, -4)$ meets the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,then the equation of the line is

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